Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 5} = \dfrac{25}{x - 5}$
Explanation: Multiply both sides by $x - 5$ $ \dfrac{x^2}{x - 5} (x - 5) = \dfrac{25}{x - 5} (x - 5)$ $ x^2 = 25$ Subtract $25$ from both sides: $ x^2 - (25) = 25 - (25)$ $ x^2 - 25 = 0$ Factor the expression: $ (x + 5)(x - 5) = 0$ Therefore $x = -5$ or $x = 5$ At $x = 5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 5$, it is an extraneous solution.